[Magical]

 To a 1.00 litre solution of 0.01M Pb(NO₃)₂ was added 20.0 mmols of Na₂SO₄.
Determine the relative loss of Pb²⁺ from solution owing to precipitation of PbSO₄.

I realize I am doing this wrong and I figure it's because I'm not calculating the ionic strength correctly.

u = 1/2 ([NO₃^-](Z)² + [Na⁺](Z)² + [SO₄^2-](Z)²)

u = 1/2 ((0.02)(1-)² + (0.04)(1+)² + (0.02)(2-)²)
u = 0.07

(I was told I should be gettin 0.05... I don't know what i'm doing wrong)

If I was to obtain an ionic strength of 0.05... I would have an (alpha)_Pb²⁺ of 0.45

And using the Debye-Hückel equation, the activity coefficient was calculated to be 0.45

Then the concentration of Pb²⁺ was calculated by using the solubility constant equation

K_sp = a_Pb²⁺ x a_SO4^2-

[Pb²⁺] = K_sp/(activity coefficient²[SO₄²⁺])

[Pb²⁺] = 6.3*10^-7 / (0.45²*0.02) = 1.6*10^-4

[Borek]

How much SO₄^2- is present in the solution?

[Magical]

0.02 mol or 0.02 M since this occurs in a 1 L solution?

[Borek]

No, that's what you have missed.

[AWK]

You can obtain 0.05 when you assume the whole PbSO₄ will precipitate (then you have 0.02 M NaNO₃ and 0.01 M Na₂SO₄ in solution).For further calculation you should take into account the ionic strength and common ion effect (sulfate) - they work in different directions.